Solved Problems - Electricity | Science
Electricity - Science
Solved Problems
1. Two bulbs are having the ratings as 60 W, 220 V and 40 W, 220 V respectively. Which one has a greater resistance?
Solution:
Electric power P is given by the formula:
$$P = \frac{V^2}{R}$$
For the same value of V, R is inversely proportional to P.
Therefore, lesser the power, greater the resistance.
Hence, the bulb with 40 W, 220 V rating has a greater resistance.
2. Calculate the current and the resistance of a 100 W, 200 V electric bulb in an electric circuit.
Solution:
Power P = 100 W and Voltage V = 200 V
Power P = V $\times$ I
3. In the circuit diagram given below, three resistors $R_1$, $R_2$ and $R_3$ of 5 Ω, 10 Ω and 20 Ω respectively are connected as shown. Calculate:
- Current through each resistor
- Total current in the circuit
- Total resistance in the circuit
Solution:
A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V).
Therefore, the current through $R_1$ is,
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B) Total current in the circuit, $I = I_1 + I_2 + I_3$
= 2 + 1 + 0.5 = 3.5 A
4. Three resistors of 1 Ω, 2 Ω and 4 Ω are connected in parallel in a circuit. If a 1 Ω resistor draws a current of 1 A, find the current through the other two resistors.
Solution:
$R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 4 Ω. Current $I_1$ = 1 A
The potential difference across the 1 Ω resistor = $I_1 R_1 = 1 \times 1 = 1 V$
Since, the resistors are connected in parallel in the circuit, the same potential difference will exist across the other resistors also.
So, the current in the 2 Ω resistor, $\frac{V}{R_2} = \frac{1}{2} = 0.5 A$
Similarly, the current in the 4 Ω resistor, $\frac{V}{R_3} = \frac{1}{4} = 0.25 A$